Abaixo está uma instrução que descreve uma tarefa de geração de SQL, combinada com uma entrada que fornece mais contexto sobre os esquemas de tabela disponíveis. Escreva o código SQL que responda adequadamente à solicitação
 ###Instrução: Quantos percentuais de pesquisa havia em outubro de 2008 quando era 30,8% em agosto de 2008?
 ###Tabelas: CREATE TABLE season_schedule (season_schedule_id AUTO_INCREMENT PRIMARY KEY, Date VARCHAR, Location VARCHAR, Opponent VARCHAR, Terps Points INT, Opp. Points INT, Record VARCHAR) CREATE TABLE liquid_measures (liquid_measures_id AUTO_INCREMENT PRIMARY KEY, Unit VARCHAR, Russian VARCHAR, Translation VARCHAR, Ratio VARCHAR, Cubic inches (exact) VARCHAR, Metric value VARCHAR, Imperial VARCHAR, US Customary VARCHAR) CREATE TABLE churches_belonging_to_the_assembly_of_canonical_bishops_of_north__&__central_america (churches_belonging_to_the_assembly_of_canonical_bishops_of_north__&__central_america_id AUTO_INCREMENT PRIMARY KEY, Adherents INT, Regular Attendees INT, % of adherents VARCHAR, Monasteries INT, Parishes INT, Average Parish Size INT) CREATE TABLE short_vowels (short_vowels_id AUTO_INCREMENT PRIMARY KEY, Short vowels VARCHAR, 17th-c. VARCHAR, American VARCHAR, British VARCHAR, Australian VARCHAR) CREATE TABLE average_polling (average_polling_id AUTO_INCREMENT PRIMARY KEY, Party VARCHAR, Result, Parliam. election 2005 VARCHAR, Result, County elections 2007 VARCHAR, Aug 2008 VARCHAR, Sep 2008 VARCHAR, Oct 2008 VARCHAR, Nov 2008 VARCHAR, Dec 2008 VARCHAR, Jan 2009 VARCHAR, Feb 2009 VARCHAR, Mar 2009 VARCHAR, Apr 2009 VARCHAR, May 2009 VARCHAR, Jun 2009 VARCHAR, Jul 2009 VARCHAR, Aug 2009 VARCHAR)  
SELECT COUNT Oct 2008 FROM average_polling WHERE Aug 2008 = 30.8%